Mesh current method

The Mesh Current Method, also known as the Loop Current Method, is quite similar to the Branch Current method in that it uses simultaneous equations, Kirchhoff's Voltage Law, and Ohm's Law to determine unknown currents in a network. It differs from the Branch Current method in that it does not use Kirchhoff's Current Law, and it is usually able to solve a circuit with less unknown variables and less simultaneous equations, which is especially nice if you're forced to solve without a calculator.
Mesh Current, conventional method

Let's see how this method works on the same example problem:

The first step in the Mesh Current method is to identify “loops” within the circuit encompassing all components. In our example circuit, the loop formed by B1, R1, and R2 will be the first while the loop formed by B2, R2, and R3 will be the second. The strangest part of the Mesh Current method is envisioning circulating currents in each of the loops. In fact, this method gets its name from the idea of these currents meshing together between loops like sets of spinning gears:

The choice of each current's direction is entirely arbitrary, just as in the Branch Current method, but the resulting equations are easier to solve if the currents are going the same direction through intersecting components (note how currents I1 and I2 are both going “up” through resistor R2, where they “mesh,” or intersect). If the assumed direction of a mesh current is wrong, the answer for that current will have a negative value.

The next step is to label all voltage drop polarities across resistors according to the assumed directions of the mesh currents. Remember that the “upstream” end of a resistor will always be negative, and the “downstream” end of a resistor positive with respect to each other, since electrons are negatively charged. The battery polarities, of course, are dictated by their symbol orientations in the diagram, and may or may not “agree” with the resistor polarities (assumed current directions):

Using Kirchhoff's Voltage Law, we can now step around each of these loops, generating equations representative of the component voltage drops and polarities. As with the Branch Current method, we will denote a resistor's voltage drop as the product of the resistance (in ohms) and its respective mesh current (that quantity being unknown at this point). Where two currents mesh together, we will write that term in the equation with resistor current being the sum of the two meshing currents.

Tracing the left loop of the circuit, starting from the upper-left corner and moving counter-clockwise (the choice of starting points and directions is ultimately irrelevant), counting polarity as if we had a voltmeter in hand, red lead on the point ahead and black lead on the point behind, we get this equation:

Notice that the middle term of the equation uses the sum of mesh currents I1 and I2 as the current through resistor R2. This is because mesh currents I1 and I2 are going the same direction through R2, and thus complement each other. Distributing the coefficient of 2 to the I1 and I2 terms, and then combining I1 terms in the equation, we can simplify as such:

At this time we have one equation with two unknowns. To be able to solve for two unknown mesh currents, we must have two equations. If we trace the other loop of the circuit, we can obtain another KVL equation and have enough data to solve for the two currents. Creature of habit that I am, I'll start at the upper-left hand corner of the right loop and trace counter-clockwise:

Simplifying the equation as before, we end up with:

Now, with two equations, we can use one of several methods to mathematically solve for the unknown currents I1 and I2:


Knowing that these solutions are values for mesh currents, not branch currents, we must go back to our diagram to see how they fit together to give currents through all components:

The solution of -1 amp for I2 means that our initially assumed direction of current was incorrect. In actuality, I2 is flowing in a counter-clockwise direction at a value of (positive) 1 amp:

This change of current direction from what was first assumed will alter the polarity of the voltage drops across R2 and R3 due to current I2. From here, we can say that the current through R1 is 5 amps, with the voltage drop across R1 being the product of current and resistance (E=IR), 20 volts (positive on the left and negative on the right). Also, we can safely say that the current through R3 is 1 amp, with a voltage drop of 1 volt (E=IR), positive on the left and negative on the right. But what is happening at R2?

Mesh current I1 is going “up” through R2, while mesh current I2 is going “down” through R2. To determine the actual current through R2, we must see how mesh currents I1 and I2 interact (in this case they're in opposition), and algebraically add them to arrive at a final value. Since I1 is going “up” at 5 amps, and I2 is going “down” at 1 amp, the real current through R2 must be a value of 4 amps, going “up:”

A current of 4 amps through R2's resistance of 2 Ω gives us a voltage drop of 8 volts (E=IR), positive on the top and negative on the bottom.

The primary advantage of Mesh Current analysis is that it generally allows for the solution of a large network with fewer unknown values and fewer simultaneous equations. Our example problem took three equations to solve the Branch Current method and only two equations using the Mesh Current method. This advantage is much greater as networks increase in complexity:

To solve this network using Branch Currents, we'd have to establish five variables to account for each and every unique current in the circuit (I1 through I5). This would require five equations for solution, in the form of two KCL equations and three KVL equations (two equations for KCL at the nodes, and three equations for KVL in each loop):

I suppose if you have nothing better to do with your time than to solve for five unknown variables with five equations, you might not mind using the Branch Current method of analysis for this circuit. For those of us who have better things to do with our time, the Mesh Current method is a whole lot easier, requiring only three unknowns and three equations to solve:

Less equations to work with is a decided advantage, especially when performing simultaneous equation solution by hand (without a calculator).

Another type of circuit that lends itself well to Mesh Current is the unbalanced Wheatstone Bridge. Take this circuit, for example:

Since the ratios of R1/R4 and R2/R5 are unequal, we know that there will be voltage across resistor R3, and some amount of current through it. As discussed at the beginning of this chapter, this type of circuit is irreducible by normal series-parallel analysis, and may only be analyzed by some other method.

We could apply the Branch Current method to this circuit, but it would require six currents (I1 through I6), leading to a very large set of simultaneous equations to solve. Using the Mesh Current method, though, we may solve for all currents and voltages with much fewer variables.

The first step in the Mesh Current method is to draw just enough mesh currents to account for all components in the circuit. Looking at our bridge circuit, it should be obvious where to place two of these currents:

The directions of these mesh currents, of course, is arbitrary. However, two mesh currents is not enough in this circuit, because neither I1 nor I2 goes through the battery. So, we must add a third mesh current, I3:

Here, I have chosen I3 to loop from the bottom side of the battery, through R4, through R1, and back to the top side of the battery. This is not the only path I could have chosen for I3, but it seems the simplest.

Now, we must label the resistor voltage drop polarities, following each of the assumed currents' directions:

Notice something very important here: at resistor R4, the polarities for the respective mesh currents do not agree. This is because those mesh currents (I2 and I3) are going through R4 in different directions. This does not preclude the use of the Mesh Current method of analysis, but it does complicate it a bit. Though later, we will show how to avoid the R4 current clash. (See Example below)

Generating a KVL equation for the top loop of the bridge, starting from the top node and tracing in a clockwise direction:


In this equation, we represent the common directions of currents by their sums through common resistors. For example, resistor R3, with a value of 100 Ω, has its voltage drop represented in the above KVL equation by the expression 100(I1 + I2), since both currents I1 and I2 go through R3 from right to left. The same may be said for resistor R1, with its voltage drop expression shown as 150(I1 + I3), since both I1 and I3 go from bottom to top through that resistor, and thus work together to generate its voltage drop.

Generating a KVL equation for the bottom loop of the bridge will not be so easy, since we have two currents going against each other through resistor R4. Here is how I do it (starting at the right-hand node, and tracing counter-clockwise):

Note how the second term in the equation's original form has resistor R4's value of 300 Ω multiplied by the difference between I2 and I3 (I2 - I3). This is how we represent the combined effect of two mesh currents going in opposite directions through the same component. Choosing the appropriate mathematical signs is very important here: 300(I2 - I3) does not mean the same thing as 300(I3 - I2). I chose to write 300(I2 - I3) because I was thinking first of I2's effect (creating a positive voltage drop, measuring with an imaginary voltmeter across R4, red lead on the bottom and black lead on the top), and secondarily of I3's effect (creating a negative voltage drop, red lead on the bottom and black lead on the top). If I had thought in terms of I3's effect first and I2's effect secondarily, holding my imaginary voltmeter leads in the same positions (red on bottom and black on top), the expression would have been -300(I3 - I2). Note that this expression is mathematically equivalent to the first one: +300(I2 - I3).

Well, that takes care of two equations, but I still need a third equation to complete my simultaneous equation set of three variables, three equations. This third equation must also include the battery's voltage, which up to this point does not appear in either two of the previous KVL equations. To generate this equation, I will trace a loop again with my imaginary voltmeter starting from the battery's bottom (negative) terminal, stepping clockwise (again, the direction in which I step is arbitrary, and does not need to be the same as the direction of the mesh current in that loop):

Solving for I1, I2, and I3 using whatever simultaneous equation method we prefer:

Example:

Use Octave to find the solution for I1, I2, and I3 from the above simplified form of equations. [octav]

Solution:

In Octave, an open source Matlab® clone, enter the coefficients into the A matrix between square brackets with column elements comma separated, and rows semicolon separated.[octav] Enter the voltages into the column vector: b. The unknown currents: I1, I2, and I3 are calculated by the command: x=A\b. These are contained within the x column vector.


octave:1>A = [300,100,150;100,650,-300;-150,300,-450]
A =
300 100 150
100 650 -300
-150 300 -450

octave:2> b = [0;0;-24]
b =
0
0
-24

octave:3> x = A\b
x =
-0.093793
0.077241
0.136092

The negative value arrived at for I1 tells us that the assumed direction for that mesh current was incorrect. Thus, the actual current values through each resistor is as such:

Calculating voltage drops across each resistor:

A SPICE simulation confirms the accuracy of our voltage calculations:[spi]

unbalanced wheatstone bridge
v1 1 0
r1 1 2 150
r2 1 3 50
r3 2 3 100
r4 2 0 300
r5 3 0 250
.dc v1 24 24 1
.print dc v(1,2) v(1,3) v(3,2) v(2,0) v(3,0)
.end



v1 v(1,2) v(1,3) v(3,2) v(2) v(3)
2.400E+01 6.345E+00 4.690E+00 1.655E+00 1.766E+01 1.931E+01

Example:

(a) Find a new path for current I3 that does not produce a conflicting polarity on any resistor compared to I1 or I2. R4 was the offending component. (b) Find values for I1, I2, and I3. (c) Find the five resistor currents and compare to the previous values.

Solution: [dvn]

(a) Route I3 through R5, R3 and R1 as shown:


Note that the conflicting polarity on R4 has been removed. Moreover, none of the other resistors have conflicting polarities.

(b) Octave, an open source (free) matlab clone, yields a mesh current vector at “x”:[octav]

octave:1> A = [300,100,250;100,650,350;-250,-350,-500]
A =
300 100 250
100 650 350
-250 -350 -500

octave:2> b = [0;0;-24]
b =
0
0
-24

octave:3> x = A\b
x =
-0.093793
-0.058851
0.136092

Not all currents I1, I2, and I3 are the same (I2) as the previous bridge because of different loop paths However, the resistor currents compare to the previous values:

IR1 = I1 + I3 = -93.793 ma + 136.092 ma = 42.299 ma
IR2 = I1 = -93.793 ma
IR3 = I1 + I2 + I3 = -93.793 ma -58.851 ma + 136.092 ma = -16.552 ma
IR4 = I2 = -58.851 ma
IR5 = I2 + I3 = -58.851 ma + 136.092 ma = 77.241 ma

Since the resistor currents are the same as the previous values, the resistor voltages will be identical and need not be calculated again.